Question: Evaluate $~~\int^\pi_0x\sin(2x) dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\pi}2$ (Choice B) B $-\dfrac{\pi}2$ (Choice C) C $0$ (Choice D) D $-\pi$ (Choice E) E $-\dfrac{\pi}3$
Answer: We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = x~$ and $~dv=\sin(2x)\, dx\,$. Then $~du = dx~$ and $~v = \int \sin(2x)\,dx = -\dfrac{\cos(2x)}{2}\,$. Integration by parts gives $ \int^\pi_0x\sin(2x)\,dx = x\cdot\Big(-\dfrac{\cos(2x)}{2}\Big)\Bigg]_0^\pi-\int^\pi_0-\dfrac{\cos(2x)}{2}dx$ $ ~~= -\dfrac{x\cos(2x)}{2}+\bigg(\dfrac{\sin(2x)}4\bigg)\Bigg]^\pi_0$ $ ~~=\Bigg(-\dfrac{\pi\,\cos(2\pi)}{2}+\dfrac{\sin(2\pi)}{4}\Bigg)$ $ -\Bigg(-\dfrac{0\cdot\,\cos\,0}{2}+\dfrac{\sin\,0}{4}\Bigg)$ $ ~~=-\dfrac{\pi}{2}+0 = -\dfrac{\pi}2$